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  哲学家进餐-多线程
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      <h1 id="哲学家进餐-多线程"><a href="#哲学家进餐-多线程" class="headerlink" title="哲学家进餐-多线程"></a>哲学家进餐-多线程</h1><h1 id="哲学家进餐"><a href="#哲学家进餐" class="headerlink" title="哲学家进餐"></a><a href="https://leetcode-cn.com/problems/the-dining-philosophers/" target="_blank" rel="noopener">哲学家进餐</a></h1><h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>5 个沉默寡言的哲学家围坐在圆桌前，每人面前一盘意面。叉子放在哲学家之间的桌面上。（5 个哲学家，5 根叉子）</p>
<p>所有的哲学家都只会在<strong>思考和进餐两种行为间交替</strong>。<strong>哲学家只有同时拿到左边和右边的叉子才能吃到面，而同一根叉子在同一时间只能被一个哲学家使用</strong>。每个哲学家吃完面后都需要把叉子放回桌面以供其他哲学家吃面。<strong>只要条件允许，哲学家可以拿起左边或者右边的叉子，但在没有同时拿到左右叉子时不能进食。</strong></p>
<p>假设面的数量没有限制，哲学家也能随便吃，不需要考虑吃不吃得下。</p>
<p><strong>设计一个进餐规则（并行算法）使得每个哲学家都不会挨饿；</strong></p>
<p>也就是说，在没有人知道别人什么时候想吃东西或思考的情况下，每个哲学家都可以在吃饭和思考之间一直交替下去。</p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201117/155006163.png" alt="mark"></p>
<a id="more"></a>



<p><strong>具体描述</strong></p>
<ol>
<li><strong>哲学家从<code>0</code> 到 <code>4</code> 按 顺时针 编号。请实现函数 <code>void wantsToEat(philosopher, pickLeftFork, pickRightFork, eat, putLeftFork, putRightFork)</code>：</strong></li>
</ol>
<ul>
<li><code>philosopher</code> 哲学家的编号。</li>
<li><code>pickLeftFork</code> 和 <code>pickRightFork</code> 表示拿起左边或右边的叉子。</li>
<li><code>eat</code> 表示吃面。</li>
<li><code>putLeftFork</code> 和 <code>putRightFork</code> 表示放下左边或右边的叉子。</li>
</ul>
<ol start="2">
<li><p><strong>由于哲学家不是在吃面就是在想着啥时候吃面，所以思考这个方法没有对应的回调。</strong></p>
</li>
<li><p><strong>给你 5 个线程，每个都代表一个哲学家，请你使用类的同一个对象来模拟这个过程。在最后一次调用结束之前，可能会为同一个哲学家多次调用该函数。</strong></p>
</li>
</ol>
<p><strong>示例</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line">输入：n &#x3D; 1</span><br><span class="line">输出：[[4,2,1],[4,1,1],[0,1,1],[2,2,1],[2,1,1],[2,0,3],[2,1,2],[2,2,2],[4,0,3],[4,1,2],[0,2,1],[4,2,2],[3,2,1],[3,1,1],[0,0,3],[0,1,2],[0,2,2],[1,2,1],[1,1,1],[3,0,3],[3,1,2],[3,2,2],[1,0,3],[1,1,2],[1,2,2]]</span><br><span class="line">解释:</span><br><span class="line">n 表示每个哲学家需要进餐的次数。</span><br><span class="line">输出数组描述了叉子的控制和进餐的调用，它的格式如下：</span><br><span class="line">output[i] &#x3D; [a, b, c] (3个整数)</span><br><span class="line">- a 哲学家编号。</span><br><span class="line">- b 指定叉子：&#123;1 : 左边, 2 : 右边&#125;.</span><br><span class="line">- c 指定行为：&#123;1 : 拿起, 2 : 放下, 3 : 吃面&#125;。</span><br><span class="line">如 [4,2,1] 表示 4 号哲学家拿起了右边的叉子。</span><br></pre></td></tr></table></figure>



<h2 id="前言"><a href="#前言" class="headerlink" title="前言"></a>前言</h2><ul>
<li>这道题本质上其实是想考察<strong>如何避免死锁</strong>。</li>
<li>易知：当 5 个哲学家都拿着其左边(或右边)的叉子时，会进入死锁。</li>
</ul>
<p><strong>PS：死锁的 4 个必要条件：</strong></p>
<ol>
<li><strong>互斥条件：</strong>一个资源每次只能被一个进程使用，即在一段时间内某 资源仅为一个进程所占有。此时若有其他进程请求该资源，则请求进程只能等待。</li>
<li><strong>请求与保持条件：</strong>进程已经保持了至少一个资源，但又提出了新的资源请求，而该资源 已被其他进程占有，此时请求进程被阻塞，但对自己已获得的资源保持不放。</li>
<li><strong>不可剥夺条件:</strong>进程所获得的资源在未使用完毕之前，不能被其他进程强行夺走，即只能 由获得该资源的进程自己来释放（只能是主动释放)。</li>
<li><strong>循环等待条件:</strong> 若干进程间形成首尾相接循环等待资源的关系。</li>
</ol>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201117/155916028.png" alt="mark"></p>
<ul>
<li><strong>情形一：允许四个哲学家进餐</strong></li>
</ul>
<p>故最多只允许 4 个哲学家去持有叉子，可保证至少有 1 个哲学家能吃上意大利面（即获得到 2个叉子）。<br>因为最差情况下是：4 个哲学家都各自持有1个叉子，此时还 剩余 1 个叉子 可供使用，这 4 个哲学家中必然有1人能获取到这个 剩余的 1 个叉子，从而手持 2 个叉子，可以吃意大利面。<br><strong>即：4 个人中，1 个人有 2 个叉子，3 个人各持 1 个叉子，共计 5 个叉子。</strong></p>
<ul>
<li><strong>情形二 ： 允许三个哲学家进餐</strong></li>
</ul>
<p><strong>当然可行，3个哲学家可以先都各自持有1把叉子，此时还剩余2把叉子；</strong></p>
<p><strong>当这3个哲学家刚好都相邻</strong>(比如：编号为图中的0, 1, 2)，可能会造成只有1个哲学家能吃到意面的情况，具体而言即0号哲学家拿到了其左侧的叉子(编号为1)，</p>
<p>1号哲学家也拿到了其左侧的叉子(编号为2)</p>
<p>2号哲学家也拿到了其左侧的叉子(编号为3)，此时只有0号哲学家能拿到其右侧的叉子(编号为0)，因此只有0号哲学家能吃到意面。<br>而其余情况下，3个哲学家中都能有2人吃到意面。</p>
<p><strong>即：3 个人中，2 个人各自持有 2 个叉子，1 个人持有 1个叉子，共计 5 个叉子。</strong></p>
<p>并且仔细想想，叉子的数目是<strong>固定</strong>的(个数为<code>5</code>)，直觉上来讲<code>3</code>个人去抢<code>5</code>个叉子 比 <code>4</code>个人去抢<code>5</code>个叉子效率高。</p>
<h2 id="1-思路一-：-互斥量实现"><a href="#1-思路一-：-互斥量实现" class="headerlink" title="1. 思路一 ： 互斥量实现"></a>1. 思路一 ： 互斥量实现</h2><ul>
<li><p>用Semaphore去实现上述的限制：<code>Semaphore eatLimit = new Semaphore(4);</code><br>一共有5个叉子，视为5个ReentrantLock，并将它们全放入1个数组中。</p>
</li>
<li><p>给叉子编号 <code>0, 1, 2, 3, 4</code>（对应数组下标）。</p>
</li>
</ul>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">DiningPhilosophers</span> </span>&#123;</span><br><span class="line">    <span class="comment">// 1. 1个Fork视为1个ReentrantLock，5个叉子即5个ReentrantLock，将其都放入数组中</span></span><br><span class="line">    <span class="keyword">private</span> <span class="keyword">final</span> ReentrantLock[] lockList = &#123;<span class="keyword">new</span> ReentrantLock(),</span><br><span class="line">            <span class="keyword">new</span> ReentrantLock(),</span><br><span class="line">            <span class="keyword">new</span> ReentrantLock(),</span><br><span class="line">            <span class="keyword">new</span> ReentrantLock(),</span><br><span class="line">            <span class="keyword">new</span> ReentrantLock()&#125;;</span><br><span class="line"></span><br><span class="line"></span><br><span class="line">    <span class="comment">// 2. 限制 ： 最多只有4个哲学家持有叉子</span></span><br><span class="line">    <span class="keyword">private</span> Semaphore eatLimit = <span class="keyword">new</span> Semaphore(<span class="number">4</span>);</span><br><span class="line"></span><br><span class="line">    <span class="function"><span class="keyword">public</span> <span class="title">DiningPhilosophers</span><span class="params">()</span> </span>&#123;</span><br><span class="line">        </span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="comment">// call the run() method of any runnable to execute its code</span></span><br><span class="line">    <span class="function"><span class="keyword">public</span> <span class="keyword">void</span> <span class="title">wantsToEat</span><span class="params">(<span class="keyword">int</span> philosopher,</span></span></span><br><span class="line"><span class="function"><span class="params">                           Runnable pickLeftFork,</span></span></span><br><span class="line"><span class="function"><span class="params">                           Runnable pickRightFork,</span></span></span><br><span class="line"><span class="function"><span class="params">                           Runnable eat,</span></span></span><br><span class="line"><span class="function"><span class="params">                           Runnable putLeftFork,</span></span></span><br><span class="line"><span class="function"><span class="params">                           Runnable putRightFork)</span> <span class="keyword">throws</span> InterruptedException </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> leftFork  = (philosopher + <span class="number">1</span>) % <span class="number">5</span>;  <span class="comment">// 左边的叉子编号</span></span><br><span class="line">        <span class="keyword">int</span> rightFork = philosopher;            <span class="comment">// 右边的叉子编号</span></span><br><span class="line"></span><br><span class="line">        <span class="comment">// 让一个人进餐 进入临界区</span></span><br><span class="line">        eatLimit.acquire();  <span class="comment">// 限制人数 -1</span></span><br><span class="line">        lockList[leftFork].lock();     <span class="comment">// 拿起左边的叉子</span></span><br><span class="line">        lockList[rightFork].lock();     <span class="comment">// 拿起右边的叉子</span></span><br><span class="line"></span><br><span class="line">        pickLeftFork.run();    <span class="comment">//拿起左边的叉子 的具体执行</span></span><br><span class="line">        pickRightFork.run();    <span class="comment">//拿起右边的叉子 的具体执行</span></span><br><span class="line">        eat.run();</span><br><span class="line"></span><br><span class="line">        putLeftFork.run();    <span class="comment">//放下左边的叉子 的具体执行</span></span><br><span class="line">        putRightFork.run();    <span class="comment">//放下右边的叉子 的具体执行</span></span><br><span class="line"></span><br><span class="line">        lockList[leftFork].unlock();    <span class="comment">//放下左边的叉子</span></span><br><span class="line">        lockList[rightFork].unlock();    <span class="comment">//放下右边的叉子</span></span><br><span class="line"></span><br><span class="line">        eatLimit.release(); <span class="comment">//限制的人数 +</span></span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>




      
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